Weak equivalences in the standard model structure on simplicial sets are allegedly closed under transfinite composition.
What's a reference for that?
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asked Jan 23, 2010 at 15:25
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3$\begingroup$ More specifics would be helpful. Do you mean that you have a transfinite composite of maps which are weak equivalences and want to show that all the objects are equivalent to the colimit? The homotopy colimit? Do you mean that you have two transfinite composites with a map between them that is a levelwise weak equivalence, and you want to show the colimits are weakly equivalent? The homotopy colimits? $\endgroup$– Tyler LawsonCommented Jan 23, 2010 at 18:20
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1$\begingroup$ Sorry if this was't clear: I want to know if the cocone component on the first object of the colimiting cocone that defines the transfinite composition is a weak equivalence. That's the morphism that is the "transfinite composite" of a transfinite sequence of weak equivalences. $\endgroup$– Urs SchreiberCommented Jan 23, 2010 at 20:39
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2 Answers
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I don't have a complete reference (and like Tyler, I don't know exactly what result you want). But here are some observations:
there is a functor $\mathrm{Ex}^\infty$, which replaces a simplicial set with a weakly equivalent fibrant replacement, and which commutes with filtered colimits. (See ch. 3 of Goerss-Jardine.)
if you have a transfinite composition(s) in which all the simplicial sets are fibrant, it is straightforward to understand their behaviour with respect to weak equivalences, using the formula for simplicial homotopy groups, which gives the right homotopy groups for Kan complexes; (in particular, simplicial homotopy groups commute with filtered colimits in this setting.) [Added later:] in particular, in any transfinite composition of weak equivalences between Kan complexes, the cocone componenent of the first object (i.e., the map from the first object to the colimit of the trasnfinite sequence) has to be a weak equivalence.
[Added:] These two facts taken together imply that a "transfinite composition" of weak equivalences is a weak equivalence.
answered Jan 23, 2010 at 20:15
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$\begingroup$ Okay, thanks. So here is what I tried: we can check if we have a weak equivalence by homming into all possible fibrant objects and checking if that produces a WE each time. So I hom the colimit cocone that defines the transfinite composition into a Kan complex. The result is now a limiting cone over a diagram of fibrant objects. I want to know the the cone component over the first (or any) object is a weak equivalence. Okay, so I now hom simplicial spheres into this and then argue that... and here I am not sure. It's probably very easy, though. $\endgroup$ Commented Jan 23, 2010 at 20:43
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$\begingroup$ @Urs. I can't see how to complete your argument. If you have a transfinite tower (aka, a cocone) of weak equivalences between fibrant objects, I see no general reason for the map from the limit to be a weak equivalence. $\endgroup$ Commented Jan 23, 2010 at 20:52
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$\begingroup$ Charles, now thanks for that updated version of your reply. Nice! I didn't think of/know if that respect for filtered colimits. With that it's clear. Thanks again! $\endgroup$ Commented Jan 23, 2010 at 23:10
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1$\begingroup$ @CharlesRezk How can I see that $\mbox{Ex}^\infty$ commutes with filtered colimits? I didn't see what you were referring to in Goerss--Jardine. $\endgroup$ Commented Nov 4, 2014 at 20:11
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1$\begingroup$ @AaronMazel-Gee It is defined degreewise by functors represented by finite simplicial sets. $\endgroup$– Zhen LinCommented Oct 19, 2015 at 22:53
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This answer serves to record two explicit proofs of this fact in the literature:
Corollary 5.1 in Raptis and Rosický, “The accessibility rank of weak equivalences”, arXiv:1403.3042v2. Theory and Applications of Categories 30:19 (2015), 687—703.
Theorem 4.6 in Barnea and Schlank, “Model structures on Ind categories and the accessibility rank of weak equivalences”, arXiv:1407.1817v6.
answered Oct 19, 2015 at 16:14